I am working on a problem in my Auction Theory textbook regarding a two-player asymmetric first price auction. Assume the bidders are risk neutral. The problem statement is as follows:

Suppose that bidder $1$'s value $X_{1}$ is distributed according to $F_{1}(x) = \frac{1}{4}(x-1)^{2}$ over $[1, 3]$, and bidder $2$'s value is distributed according to $\text{exp}(\frac{2}{3}x - 2)$ over $[0, 3]$. Show that $\beta_{1}(x) = x - 1$ and $\beta_{2}(x) = \frac{2}{3}x$ constitute equilibrium bidding strategies in a first price auction.

I am trying to work on deriving $\beta_{1}$ and $\beta_{2}$. Unfortunately, my knowledge of differential equations isn't terribly strong. Would someone be able to double check my work and let me know if I have logic errors? I have derived the correct bidding functions, but am not entirely confident my work is sound.

First, suppose the equilibrium bidding functions $\beta_{1} : [1, 3] \to \mathbb{R}_{+}, \beta_{2} : [0, 3] \to \mathbb{R}_{+}$ are strictly increasing and differentiable. Define $g_{1}(x) = \beta_{1}^{-1}(x)$ and $g_{2}(x) = \beta_{2}^{-1}(x)$.

Player $i$ with valuation $v$ can only vary his bid, so he seeks to find the optimal bid given by the optimization problem below.

$$\max_{b} F_{-i}(g_{-i}(b)) \cdot (v - b)$$

We consider the First Order Conditions:

$$F_{-i}(g_{-i}(b)) = \dfrac{f_{-i}(g_{-i}(b))}{\beta_{-i}^{\prime}(g_{-i}(b))} \cdot (v-b)$$

At equilibrium, $v = g_{i}(b)$. Applying this and noting $\dfrac{1}{\beta_{-i}^{\prime}(g_{-i}(b))} = (g_{-i}(b))^{\prime}$, we have:

$$(g_{-i}(b))^{\prime} = \dfrac{F_{-i}(g_{-i}(b))}{f_{-i}(g_{-i}(b))} \cdot \dfrac{1}{g_{i}(b) - b}$$

Plugging in each $F_{i}$, we obtain:

$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}$$


$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}$$

At equilibrium, we have $\beta_{1}(3) = \beta_{2}(3)$. By individual rationality, $\beta_{2}(0) = 0 \implies g_{2}(0) = 0$.

While I could obviously use the problem statement that $\beta_{1}(x) = x - 1$ to conclude that $g_{1}(0) = 1$, I don't know how to justify this boundary condition independently. Does anyone have any insights into this?

Assuming this boundary condition though, I note:

$$g_{2}^{\prime}(0) = \dfrac{3}{2} \cdot \dfrac{1}{1 - 0} = \dfrac{3}{2}$$

From here, I can wave my hand and guess that $g_{2}^{\prime}(b) = \dfrac{3}{2}$, which would imply $g_{2}(b) = \dfrac{3}{2}b$. I'm not sure how to formally derive this though. Would anyone have insights into this?

Once I have $g_{2}(b) = \dfrac{3}{2}b$, I can plug into $g_{1}^{\prime}(b)$ to get:

$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{\dfrac{3}{2}b - b} = \dfrac{g_{1}(b) - 1}{b}$$

Which is a first order linear differential equation, whose solution is:

$g_{1}(b) = b + 1 \implies \beta_{1}(v) = v - 1$.

And we have $\beta_{2}(v) = \dfrac{2}{3}v$.

My work is certainly a little hand-wavy. I would greatly appreciate any help in solidifying the details. Thank you in advance for any help!

  • 1
    $\begingroup$ You can not have $\beta_1(0)=1$ since the lowest type of player 1 is $1$. You must have $\beta_1(1)=0$ since the eq. is non-atomic (no bids have positive probability). $\endgroup$ Aug 5, 2015 at 20:26
  • $\begingroup$ Whoops. I meant that to say $g_{1}(0) = 1$. Sorry for the confusion. :-) $\endgroup$
    – ml0105
    Aug 5, 2015 at 20:28
  • 1
    $\begingroup$ $\beta_{2}(0) = 0 \implies g_{2}(0) = 0$ is implicitly assuming the solution is strictly increasing -- if you don't want to use the info about the solutions then in general, it is possible for a player to have $\beta_2=0$ in a range of low types. The boundary conditions that we use are the ones at the top: $g_1(\overline b)=g_2(\overline b)$. Usually we find $\overline b$ later after solving the system. $\endgroup$ Aug 5, 2015 at 21:02
  • $\begingroup$ I've never encountered such a case. I've always assumed the bidding functions to be strictly increasing and differentiable. It sounds like this might not be the best practice, from what you're saying, though. $\endgroup$
    – ml0105
    Aug 5, 2015 at 21:07
  • $\begingroup$ If you have only two players and the lowest bid is zero then you will not have atoms at the lowest bid. But in the FPA it is possible for the lowest winning bid be above zero in eq. I deleted my previous comment because we can only be sure zero is the lowest bid in the All-Pay Auction and here we are talking about the FPA. $\endgroup$ Aug 5, 2015 at 21:40

2 Answers 2


Since you are given the equilibrium. You can solve for $g_1$ and $g_2$. $$ g_1(b)=b+1\quad\text{ and }\quad g_2(b)=\frac 32 b$$ It is then straightforward to show that these $g_1$ and $g_2$ solve the ODE system:

\begin{align*} &g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}\\ &g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}\\ &g_1(0)=1\\ &g_2(0)=0.\end{align*}

Since we have independent, private values. The single-crossing condition holds in the FPA so the FOC are sufficient for a Nash=Bayes eq.

Probably, what you would like to do is to solve the above system directly without knowing what was the solution. Since the system is not autonomous, I don't know any trick but any CAS (computer algebra software): Maple or Mathematica or Sage should be able to solve it for you.

  • $\begingroup$ I am aware of that. I was hoping to be able to work through a derivation of these solutions without knowing the answer up front. Thank you for your help, regardless. $\endgroup$
    – ml0105
    Aug 5, 2015 at 20:38
  • $\begingroup$ Do you know as well why $g_{1}(0) = 1$ is a valid boundary condition without knowing the solution up front? Is there a result in auction theory stating that it is an equilibrium bidding strategy to bid $0$ if you have value $a \in [a, b]$? I'm not quite convinced of the economic intuition of this, and my differential equations/modeling skills aren't strong enough for me to justify it that way. $\endgroup$
    – ml0105
    Aug 5, 2015 at 20:47

I followed someone's suggestion here, which was quite fruitful. So we have the differential equations:

$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}$$


$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}$$

With the boundary conditions $g_{2}(0) = 0$ and $g_{1}(\overline{b}) = g_{2}(\overline{b}) = 3$, where $\overline{b}$ is the maximum bid.

Now we guess that $g_{1}(b) = \alpha b + \gamma$ and $g_{2}(b) = \delta b + \lambda$. Applying $g_{2}(b) = 0$ yields that $\lambda = 0$.

Next, I substitute $g_{1}(b)$ into $g_{2}^{\prime}(b)$ to obtain:

$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{(\alpha - 1)b + \gamma}$$

Integrating $g_{2}^{\prime}(b)$ yields

$$g_{2}(b) = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)b + \gamma)$$

We note there is no constant when integrating, as $g_{2}(b) = \delta b$. We now apply $g_{2}(0) = 0$ again, concluding that $ln( \gamma) = 0$. And so $\gamma = 1$. Thus, $g_{1}(b) = \alpha b + 1$.

We now solve:

$$g_{2}(\overline{b}) = 3 = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)\overline{b} + \gamma)$$

From this and noting that $g_{1}(\overline{b}) = 3 = \alpha \overline{b} + 1$, we obtain:

$$e^{2(\alpha - 1)} = 3 - \overline{b} \implies \overline{b} = 3 - e^{2(\alpha - 1)}$$

Plugging this into $g_{1}(b)$ yields:

$$g_{1}(\overline{b}) = \alpha(3 - e^{2(\alpha - 1)}) + 1 = 3$$

Which implies that the solution $\alpha = 1$. Thus, $\overline{b} = 2$.

So $\delta = \dfrac{3}{2}$.

Thus, $g_{1}(b) = b + 1 \implies \beta_{1}(v) = v - 1$; and $g_{2}(b) = \dfrac{3}{2}b$ implies $\beta_{2}(v) = \dfrac{2}{3}v$ as desired.

If anyone could clarify a reason for guessing at $g_{1}(b)$ and $g_{2}(b)$ to be affine without knowing the solution, that would be greatly appreciated. Otherwise, with the guess, the solution was quite a bit stronger than my previous attempt.

  • $\begingroup$ There is no reason for the inverse bids to be affine. You are just lucky this is the case here. $\endgroup$ Aug 7, 2015 at 18:33

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